# inequality involving heights and bisectors

Let $a,b,c,a \le b \le c$ be the sides of the triangle $ABC$, $l_a,l_b,l_c$ the lengths of its bisectors and $h_a,h_b,h_c$ the lengths of its heights. Prove that: $$\frac {h_a+h_c} {h_b} \ge \frac {l_a+l_c} {l_b}$$

• Just to make sure: you are referring to angle bisectors, right? (Not bisectors of the opposite side) Apr 30, 2016 at 10:09
• angle bisectors @Andreas Apr 30, 2016 at 10:32

The formulae for the lenghts of the angular bisectors and the heights are known. (see e.g. https://en.wikipedia.org/wiki/Triangle ). Let $T$ be the area of the triangle. We have $$l_a = \sqrt{bc (1 - \frac{a^2}{(b+c)^2})}$$ and $$h_a = \frac{2 T}{a}$$ and cyclic shifts of those.
With these formulae, the required inequality gets $$\frac{1/a + 1/c}{1/b} \geq \frac{\sqrt{bc (1 - \frac{a^2}{(b+c)^2})} + \sqrt{ab (1 - \frac{c^2}{(a+b)^2})}}{\sqrt{ac (1 - \frac{b^2}{(a+c)^2})}}$$ Multiplying the RHS by $$1 = \frac{1/\sqrt{abc}}{1/\sqrt{abc}}$$ gives $$\frac{1/a + 1/c}{1/b} \geq \frac{\frac{1}{a}\sqrt{a (1 - \frac{a^2}{(b+c)^2})} + \frac{1}{c}\sqrt{c(1 - \frac{c^2}{(a+b)^2})}}{\frac{1}{b}\sqrt{b (1 - \frac{b^2}{(a+c)^2})}}$$ Due to homogeneity, we can set $b=1$. This gives the condition $a\leq 1\leq c$ and the inequality gets
$$\frac{1}{a} + \frac{1}{c} \geq \frac{1}{a}\sqrt{ a \frac{ 1 - \frac{a^2}{(1+c)^2} }{ 1 - \frac{1}{(a+c)^2} } } + \frac{1}{c}\sqrt{ c\frac{ 1 - \frac{c^2}{(a+1)^2} }{ 1 - \frac{1}{(a+c)^2} } }$$ We now first prove that the first root is less or equal than 1, i.e. we need $${a (1 - \frac{a^2}{(1+c)^2})} \leq 1 - \frac{1}{(a+c)^2}$$ which can be transformed into $$(a+c)^2 (1-a) \geq 1 - a^3 (\frac{a+c}{1+c})^2$$ The last bracket can be expanded and it suffices to prove $$(a+c)^2 (1-a) \geq 1 - a^3 (1 - 2 \frac{1-a}{1+c})$$ which shortens to $$(1+c) a (1-a) \geq 2 a^3 (1-a)$$ or $$(1+c) \geq 2 a^2$$ which is certainly true under the given condition $a\leq 1\leq c$.
Likewise, we prove that the second root is less or equal than 1, i.e. we need $${c (1 - \frac{c^2}{(1+a)^2})} \leq 1 - \frac{1}{(a+c)^2}$$ which can be transformed into $$(a+c)^2 (1-c) \geq 1 - c^3 (\frac{a+c}{1+a})^2$$ We can expand the last bracket and it suffices to prove $$(a+c)^2 (1-c) \geq 1 - c^3 (1 + 2 \frac{c-1}{1+a})$$ or in positive terms $$(a+c)^2 (c-1) \leq c^3 -1 + 2 c^3 \frac{c-1}{1+a}$$ Since $a\leq1$, it suffices to prove $$(1+c)^2 (c-1) - c^3 +1 \leq 2 c^3 \frac{c-1}{1+a}$$ or $$(1+a)(c^2 - c) \leq 2 c^2 (c^2-c)$$ which shortens to $$1+a \leq 2 c^2$$ which is certainly true under the given condition $a\leq 1\leq c$.